\(\int \frac {1}{x^2 \sqrt {1-x^4}} \, dx\) [892]

Optimal result

Integrand size = 15, antiderivative size = 27 \[ \int \frac {1}{x^2 \sqrt {1-x^4}} \, dx=-\frac {\sqrt {1-x^4}}{x}-E(\arcsin (x)|-1)+\operatorname {EllipticF}(\arcsin (x),-1) \]

[Out]

-EllipticE(x,I)+EllipticF(x,I)-(-x^4+1)^(1/2)/x

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {331, 313, 227, 1195, 435} \[ \int \frac {1}{x^2 \sqrt {1-x^4}} \, dx=\operatorname {EllipticF}(\arcsin (x),-1)-E(\arcsin (x)|-1)-\frac {\sqrt {1-x^4}}{x} \]

[In]

Int[1/(x^2*Sqrt[1 - x^4]),x]

[Out]

-(Sqrt[1 - x^4]/x) - EllipticE[ArcSin[x], -1] + EllipticF[ArcSin[x], -1]

Rule 227

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[EllipticF[ArcSin[Rt[-b, 4]*(x/Rt[a, 4])], -1]/(Rt[a, 4]*Rt[
-b, 4]), x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]

Rule 313

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[-b/a, 2]}, Dist[-q^(-1), Int[1/Sqrt[a + b*x^4]
, x], x] + Dist[1/q, Int[(1 + q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && NegQ[b/a]

Rule 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 435

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*Ell
ipticE[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d))], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 0
]

Rule 1195

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[(-a)*c, 2]}, Dist[Sqrt[-c], Int
[(d + e*x^2)/(Sqrt[q + c*x^2]*Sqrt[q - c*x^2]), x], x]] /; FreeQ[{a, c, d, e}, x] && GtQ[a, 0] && LtQ[c, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {\sqrt {1-x^4}}{x}-\int \frac {x^2}{\sqrt {1-x^4}} \, dx \\ & = -\frac {\sqrt {1-x^4}}{x}+\int \frac {1}{\sqrt {1-x^4}} \, dx-\int \frac {1+x^2}{\sqrt {1-x^4}} \, dx \\ & = -\frac {\sqrt {1-x^4}}{x}+F\left (\left .\sin ^{-1}(x)\right |-1\right )-\int \frac {\sqrt {1+x^2}}{\sqrt {1-x^2}} \, dx \\ & = -\frac {\sqrt {1-x^4}}{x}-E\left (\left .\sin ^{-1}(x)\right |-1\right )+F\left (\left .\sin ^{-1}(x)\right |-1\right ) \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.67 \[ \int \frac {1}{x^2 \sqrt {1-x^4}} \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},x^4\right )}{x} \]

[In]

Integrate[1/(x^2*Sqrt[1 - x^4]),x]

[Out]

-(Hypergeometric2F1[-1/4, 1/2, 3/4, x^4]/x)

Maple [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4.

Time = 4.31 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.56

method	result	size
meijerg	\(-\frac {{}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};x^{4}\right )}{x}\)	\(15\)
default	\(-\frac {\sqrt {-x^{4}+1}}{x}+\frac {\sqrt {-x^{2}+1}\, \sqrt {x^{2}+1}\, \left (F\left (x , i\right )-E\left (x , i\right )\right )}{\sqrt {-x^{4}+1}}\)	\(53\)
elliptic	\(-\frac {\sqrt {-x^{4}+1}}{x}+\frac {\sqrt {-x^{2}+1}\, \sqrt {x^{2}+1}\, \left (F\left (x , i\right )-E\left (x , i\right )\right )}{\sqrt {-x^{4}+1}}\)	\(53\)
risch	\(\frac {x^{4}-1}{x \sqrt {-x^{4}+1}}+\frac {\sqrt {-x^{2}+1}\, \sqrt {x^{2}+1}\, \left (F\left (x , i\right )-E\left (x , i\right )\right )}{\sqrt {-x^{4}+1}}\)	\(57\)

[In]

int(1/x^2/(-x^4+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/x*hypergeom([-1/4,1/2],[3/4],x^4)

Fricas [A] (verification not implemented)

none

Time = 0.08 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {1}{x^2 \sqrt {1-x^4}} \, dx=-\frac {x E(\arcsin \left (x\right )\,|\,-1) - x F(\arcsin \left (x\right )\,|\,-1) + \sqrt {-x^{4} + 1}}{x} \]

[In]

integrate(1/x^2/(-x^4+1)^(1/2),x, algorithm="fricas")

[Out]

-(x*elliptic_e(arcsin(x), -1) - x*elliptic_f(arcsin(x), -1) + sqrt(-x^4 + 1))/x

Sympy [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 32 vs. \(2 (15) = 30\).

Time = 0.39 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {1}{x^2 \sqrt {1-x^4}} \, dx=\frac {\Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{2} \\ \frac {3}{4} \end {matrix}\middle | {x^{4} e^{2 i \pi }} \right )}}{4 x \Gamma \left (\frac {3}{4}\right )} \]

[In]

integrate(1/x**2/(-x**4+1)**(1/2),x)

[Out]

gamma(-1/4)*hyper((-1/4, 1/2), (3/4,), x**4*exp_polar(2*I*pi))/(4*x*gamma(3/4))

Maxima [F]

\[ \int \frac {1}{x^2 \sqrt {1-x^4}} \, dx=\int { \frac {1}{\sqrt {-x^{4} + 1} x^{2}} \,d x } \]

[In]

integrate(1/x^2/(-x^4+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(-x^4 + 1)*x^2), x)

Giac [F]

\[ \int \frac {1}{x^2 \sqrt {1-x^4}} \, dx=\int { \frac {1}{\sqrt {-x^{4} + 1} x^{2}} \,d x } \]

[In]

integrate(1/x^2/(-x^4+1)^(1/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(-x^4 + 1)*x^2), x)

Mupad [B] (verification not implemented)

Time = 5.69 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.48 \[ \int \frac {1}{x^2 \sqrt {1-x^4}} \, dx=-\frac {{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ x^4\right )}{x} \]

[In]

int(1/(x^2*(1 - x^4)^(1/2)),x)

[Out]

-hypergeom([-1/4, 1/2], 3/4, x^4)/x